Question

two point charges +2nC and -2nC are placed apart 0.1 m apart .Find the resultant electric intensity at a point at distance 0.1m from each charge.

Answer 1

Refer the following attachment for the answer and the steps used to obtain it.

Answer 2

The magnitude of resultant electric field is $1.8\times10^{3}\ N/C$

Explanation:

Given that,

First charge = 2 nC

Second charge = -2 nC

Distance = 0.1 m

We need to calculate the electric field for first charge

Using formula of electric field

$E_{1}=\dfrac{kq_{1}}{r^2}$

Put the value into the formula

$E_{1}= \dfrac{9\times10^{9}\times2\times10^{-9}}{(0.1)^2}$

$E_{1}=1800\ N/C$

$E_{1}=1.8\times10^{3}\ N/C$

We need to calculate the electric field for second charge

Using formula of electric field

$E_{2}=\dfrac{kq_{2}}{r^2}$

Put the value into the formula

$E_{2}= \dfrac{9\times10^{9}\times2\times10^{-9}}{(0.1)^2}$

$E_{2}=1800\ N/C$

$E_{2}=1.8\times10^{3}\ N/C$

Here,

$E_{1}=E_{2}$

We need to calculate the resultant electric field

Using formula of electric field

$E_{R}=\sqrt{E_{1}^2+E_{2}^2+2E_{1}E_{2}\cos\theta}$

Here, $\theta=120^{\circ}$

Put the value into the formula

$E_{R}=\sqrt{E^2+E^2-2\dfrac{E^2}{2}}$

$E_{R}=E$

$E_{R}=1.8\times10^{3}\ N/C$

Hence, The magnitude of resultant electric field is $1.8\times10^{3}\ N/C$

Learn more :    

Topic : electric field

https://brainly.in/question/3244061